MaxWong

btw that's \ddagger not \dagger

All except \infty from http://omega.albany.edu:8008/Symbols.html

I discovered \infty myself.

Do the brackets first.

8 = 6x + 42 + x

Then combine like terms

8 = 7x+42

Move the 42 to the other side of the equal sign,

7x = 8 - 42 = -34

x = \(\color{blue }-\dfrac{34}{7}\)

No one can see your question so I can't help you. Joking. Just highlight it.

Using the sine rule, \(\color{blue}\dfrac{a}{sin48}=\dfrac{5}{sin26} \\ \color{blue}a=\frac{5sin48}{sin26} \approx 8.47\)

\(\S\)

\(\S \infty \S\)

\(\S\infty\infty\infty\S\)

\(\S\infty\S\)

\(\S\)

\(\color{aqua}sec^2\theta + csc^2\theta=\color{aqua}\frac{1}{cos^2\theta}+\frac{1}{sin^2\theta}=\frac{sin^2\theta+cos^2\theta}{sin^2\theta cos^2\theta}= \frac{1}{sin^2\theta cos^2\theta}\space OR\space sec^2\theta csc^2\theta\)

(Same as CPhill's answer, just using LaTeX to make it clearer.)

:D

\(\frac{1.0\times10^{-14}}{7.92\times 10^5}\)

\(=\frac{1}{7.92}\times \frac{1}{10^{19}}\approx 0.1262626\times 10^{-19}\)

32 = 7-5y

y = \(\frac{7-32}{5}=-5\)

I can't comprehend this question. Please present the question in a better way.

log 53

0| 1 1 1 2 2 2 4 8 8

1| 9

2| 3 6 7